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#1 2008-11-12 14:01:45

Daniel123
Member
Registered: 2007-05-23
Posts: 663

Quick question

I'm almost there with this one... I just can't make the final leap.

"Prove that if x,y,z are integers such that 9|(x^2+y^2+z^2), then 9|(x^2-y^2) or 9|(y^2-z^2) or 9|(z^2-x^2)."

The quadratic residues (mod 9) are 1, 4 and 7. Therefore the only ways that three squares can be summed to give 0 (mod 9) are (0,0,0); (1,1,7); (1,4,4); and (4,7,7) and any permutations of these triples. In all cases there is a repeat, but does this / why does this mean that 9|(x^2-y^2) or 9|(y^2-z^2) or 9|(z^2-x^2)?

I can sort of see it, but I can't explain it.

Thanks smile

Last edited by Daniel123 (2008-11-12 14:02:03)

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#2 2008-11-12 20:39:38

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Quick question

The fact that there's always a repeat is the key thing.
This means that there are always a and b in the triple so that a^2 = b^2 (mod 9).
So a^2 - b^2 = 0 (mod 9), ie. 9| (a^2 - b^2).

Now prove that {a,b} has to be one of {x,y}; {x,z} and {y,z} and you're there.


Why did the vector cross the road?
It wanted to be normal.

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#3 2008-11-13 03:46:01

Daniel123
Member
Registered: 2007-05-23
Posts: 663

Re: Quick question

smile thanks.

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